Solve the equations.`tan^(-1)((1-x)/(1+x))=1/2tan^(-1)x ,(x >0)`

Explanation

To solve the equation \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1}(x) \) for \( x > 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1}(x) \]

### Step 2: Use the double angle formula for tangent
We know that:
\[ 2 \tan^{-1}(y) = \tan^{-1}\left(\frac{2y}{1-y^2}\right) \]
Applying this to the right side, we can rewrite:
\[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}\left(\frac{x}{2}\right) \]

### Step 3: Set the arguments equal
Since the tangent function is one-to-one in the range of the inverse tangent, we can equate the arguments:
\[ \frac{1-x}{1+x} = \frac{x}{2} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 2(1-x) = x(1+x) \]

### Step 5: Expand both sides
Expanding both sides results in:
\[ 2 - 2x = x + x^2 \]

### Step 6: Rearrange the equation
Rearranging the equation gives:
\[ x^2 + 3x - 2 = 0 \]

### Step 7: Factor the quadratic equation
We can factor the quadratic:
\[ (x + 2)(x - 1) = 0 \]

### Step 8: Solve for \( x \)
Setting each factor to zero gives us:
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \quad (\text{not valid since } x > 0) \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 9: Verify the solution
Since \( x = 1 \) is greater than 0, it is a valid solution.

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{1} \]